\section{How Order 2 Tensor's Invariants Come From}
A brief proof for why order 2 tensor's invariants are invariant under rotation of basis.

To make the proof brief, I will just show order 2 rank 1 case, which implies that I can represent my order 2 tensor as a tensor product of two order 1 tensors,

\begin{equation}\nonumber
\textbf{A} = \textbf{ab}^T
\end{equation}

For the original basis, we can get tensor \textbf{A}'s invariants from the characteristic polynomial,

\begin{equation}\nonumber
\textit{det}(\textbf{A} - \lambda\textbf{I}) = 0
\end{equation}

By applying arbitary orthogonal transoformation \textbf{Q} to the basis, the tensor becomes

\begin{equation}\nonumber
\textbf{A}^{\ast} = \textbf{Qab}^T\textbf{Q}^T = \textbf{QAQ}^T
\end{equation}

The characteristic polynomial for the new tensor becomes

\begin{equation}\nonumber
\begin{split}
&\textit{det}(\textbf{QAQ}^T - \lambda^{\ast}\textbf{I}) = 0\\
\Rightarrow &\textit{det}(\textbf{QAQ}^T - \lambda^{\ast}\textbf{QIQ}^T) = 0\\
\Rightarrow &\textit{det}(\textbf{Q})\textit{det}(\textbf{A} - \lambda^{\ast}\textbf{I})\textit{det}(\textbf{Q}^T) = 0\\
\Rightarrow &\textit{det}(\textbf{A} - \lambda^{\ast}\textbf{I}) = 0
\end{split}
\end{equation}